GATE 2026 EE Q.45: solution and answer

GATE 2026 · EEQ.45 · 2 marks · MCQ

The digital circuit shown has 3 inputs ( $x$ , $y$ , and $z$ ).

[Figure]

The simplified logical expression for the output (OUT) is:

GATE 2026 EE question 45: Three inputs x, y, z enter from the left on three horizontal lines, x on top, then y, then z. — Figure from the GATE 2026 EE paper. It is part of the question.

(A)
$\overset{x}{ˉ} \overline{y z}$
(B)
$0$
(C)
$\overset{x}{ˉ} (y + z)$
(D)
$1$

Answer and solution

ANSWER

OUT $= \overline{x + y + z} \cdot \overset{x}{ˉ} (y + z) = 0$ for every input, so the answer is (B) $0$ .

Official answer: BSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Notation: an overbar means complement (so $\overset{x}{ˉ}$ is NOT $x$ ). Read each gate from the junction dots. A jump (a small arc) where two wires cross means they are NOT connected.
NOR gate, inputs $x, y, z$ : $N = \overline{x + y + z} = \overset{x}{ˉ} \overset{y}{ˉ} \overset{z}{ˉ}$ (De Morgan's theorem).
NOT gate, input tapped from the $x$ line: its output is $\overset{x}{ˉ}$ . The junction dot on that output sends $\overset{x}{ˉ}$ to both AND gates.
AND gate 1 has inputs $\overset{x}{ˉ}$ and $y$ , so its output is $\overset{x}{ˉ} y$ . AND gate 2 has inputs $\overset{x}{ˉ}$ and $z$ , so its output is $\overset{x}{ˉ} z$ .
OR gate: $P = \overset{x}{ˉ} y + \overset{x}{ˉ} z = \overset{x}{ˉ} (y + z)$ (take out the common factor with the distributive law).
Final AND gate: $OUT = N \cdot P = \overset{x}{ˉ} \overset{y}{ˉ} \overset{z}{ˉ} \cdot \overset{x}{ˉ} (y + z) = \overset{x}{ˉ} \overset{y}{ˉ} \overset{z}{ˉ} (y + z) (\overset{x}{ˉ} \overset{x}{ˉ} = \overset{x}{ˉ}) = \overset{x}{ˉ} (\overset{y}{ˉ} y \overset{z}{ˉ} + \overset{y}{ˉ} \overset{z}{ˉ} z) = \overset{x}{ˉ} (0 + 0) = 0$ This uses $y \overset{y}{ˉ} = 0$ and $z \overset{z}{ˉ} = 0$ .
Check: $N = 1$ only when $x = y = z = 0$ . For that input $P = \overset{x}{ˉ} (y + z) = 1 \cdot 0 = 0$ . So the two inputs of the final AND gate are never both 1, and OUT is 0 for every input combination. Answer: (B).

KEY IDEA

The NOR output $\overset{x}{ˉ} \overset{y}{ˉ} \overset{z}{ˉ}$ is 1 only when $y = z = 0$ . The OR output $\overset{x}{ˉ} (y + z)$ needs $y$ or $z$ to be 1. The final AND gate needs both to be 1 at once, which never happens, so OUT is the constant 0.

WATCH OUT FOR

Stopping at the OR gate and choosing (C) $\overset{x}{ˉ} (y + z)$ , even though that signal still passes through the final AND gate with the NOR output. A second mistake is misusing De Morgan and writing $\overline{x + y + z}$ as $\overset{x}{ˉ} \overline{y z}$ (option A); the correct form is $\overset{x}{ˉ} \overset{y}{ˉ} \overset{z}{ˉ}$ . A third is treating the wire jumps as connections, which puts the wrong input on the AND gates.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 22 Boolean logic questions, with notes

GATE 2026 EE Q.45: solution and answer

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